Parametric Equations for Line of Intersection of 3x-6y-2z=15 & 2x+y-2z=5

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Homework Help Overview

The discussion revolves around finding parametric equations for the line of intersection between two planes defined by the equations 3x − 6y − 2z = 15 and 2x + y − 2z = 5.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various attempts to derive the parametric equations, with one participant questioning the validity of a proposed solution and suggesting verification against the original plane equations. Another participant provides a different approach involving cross products and setting a variable to zero.

Discussion Status

The discussion includes multiple attempts at finding the correct parametric equations, with some participants verifying their results against the plane equations. There is an ongoing exploration of different methods, but no explicit consensus has been reached regarding the final solution.

Contextual Notes

Participants are encouraged to check their solutions against the original equations, indicating a focus on accuracy and verification in the problem-solving process.

helpm3pl3ase
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Homework Statement



Find parametric equations for the line in which the planes 3x − 6y − 2z = 15
and 2x + y − 2z = 5 intersect.


Homework Equations





The Attempt at a Solution



<2, 1, -2> - <3, -6, -2> = <-1, 7, 0>

x = 2 - t, y = 1 + 7t, z = -2

Did I do this correctly??
 
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Uh, no. Put the solution back into your two line equations. Does it work? I don't think so. Where did you find this 'solution method'? I think you should maybe try and find one that does.
 
<3, -6, -2> X <2, 1, -2> = <14, 2, 15>

Then set z to 0 to get x = 3, y = -1 ==> <3, -1, 0>

x = 3 + 14t
y = -1 + 2t
z = 15t
 
Better. Did you check by substituting your result back into the plane equations?
 
yes it worked, got 5 and 15
 
helpm3pl3ase said:
yes it worked, got 5 and 15

Good!
 

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